Question

1. Tiger Woods is standing on the 18^th green at 2019 Masters. It is round 4 and he is winning by 2 strokes. He reads the green and realizes it’s a straight shot; the green is completely level. He putts his ball and the instant his putter makes contact with the ball (t=0), the ball has a velocity of 2 m·s.
   1. If the acceleration of the ball is constant at -0.4 m·s², how long in seconds will it take the ball to come to a complete stop (t = ?) (? (2 pts)
   2. What is the average velocity of the ball from the instant the putter makes contact (t = 0) to when the ball comes to a stop (t = ?)? (2 pts)

C. From where Tiger putt the ball to hole is 5.2 meters in length. His putt comes up just short. How much further does Tiger have to go for his second putt? (2 pts

Answer 1

a)

Using $v_f=v_i+at$

$v_i=2\,m/s ,v_f=0\,m/s,a=-0.4\,m/s^2$

$t=\frac{v_f-v_i}{a}=\frac{0-2}{-0.4}=5\,s$

Time taken for the ball to come to rest is $t=5\,s$

b)

Total displacement of  ball is $s=v_it+\frac{1}{2}at^2=2*5+\frac{1}{2}(-0.4)(5)^2=5\,m$

Average velocity of ball = Total displacement /total time

$v_{avg}=\frac{s}{t}=\frac{5}{5}=1\,m/s$

c) In the first putt, the ball moves through a distance of $5\,m$

To reach the hole, in the second putt ,the ball must travel a further distance $5.2-5=0.2\,m$